A273278 -id:A273278 - cp's OEIS frontend

A273294 Least nonnegative integer m such that there are nonnegative integers x,y,z,w for which x^2 + y^2 + z^2 + w^2 = n and x + 3y + 5z = m^2.

0, 0, 1, 2, 0, 1, 2, 3, 4, 0, 1, 2, 3, 3, 3, 4, 0, 1, 2, 3, 2, 3, 3, 4, 4, 0, 1, 2, 3, 3, 4, 4, 2, 3, 3, 4, 0, 1, 2, 3, 4, 2, 3, 6, 4, 3, 3, 6, 4, 0, 1, 2, 2, 3, 5, 4, 4, 4, 3, 4, 5, 5, 3, 4, 0, 1, 2, 3, 4, 5, 4, 6, 4, 3, 4, 4, 4, 3, 4, 4, 2

Offset: 0

Zhi-Wei Sun, May 19 2016

nonn

Clearly, a(n) = 0 if n is a square. Part (i) of the conjecture in A271518 implies that a(n) always exists.

For more conjectural refinements of Lagrange's four-square theorem, one may consult arXiv:1604.06723.

			a(1) = 0 since 1 = 0^2 + 0^2 + 0^2 + 1^2 with 0 + 3*0 + 5*0 = 0^2.
a(2) = 1 since 2 = 1^2 + 0^2 + 0^2 + 1^2 with 1 + 3*0 + 5*0 = 1^2.
a(3) = 2 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with 1 + 3*1 + 5*0 = 2^2.
a(3812) = 11 since 3812 = 37^2 + 3^2 + 15^2 + 47^2 with 37 + 3*3 + 5*15 = 11^2.
a(3840) = 16 since 3840 = 48^2 + 16^2 + 32^2 + 16^2 with 48 + 3*16 + 5*32 = 16^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Cf. A000118, A000290, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273107, A273108, A273110, A273134, A273278.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[m=0;Label[bb];Do[If[3y+5z<=m^2&&SQ[n-y^2-z^2-(m^2-3y-5z)^2],Print[n," ",m];Goto[aa]],{y,0,Sqrt[n]},{z,0,Sqrt[n-y^2]}];m=m+1;Goto[bb];Label[aa];Continue,{n,0,80}]

A273302 Least nonnegative integer x such that n = x^2 + y^2 + z^2 + w^2 for some nonnegative integer y,z,w with x + 3y + 5z a square.

Original entry on oeis.org

0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 2, 0, 0, 0, 1, 1, 2, 2, 1, 0, 0, 0, 1, 1, 0, 0, 5, 4, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 2, 4, 0, 0, 1, 1, 4, 0, 1, 0, 0, 0, 0, 5, 0, 0, 2, 0, 0, 1, 0, 1, 1, 2, 3, 0, 0, 0, 1, 0, 0, 0, 3, 4

Offset: 0

Zhi-Wei Sun, May 19 2016

nonn

Clearly, a(n) = 0 if n is a square. Part (i) of the conjecture in A271518 implies that a(n) always exists.
Compare this sequence with A273294.
For more conjectural refinements of Lagrange's four-square theorem, one may consult arXiv:1604.06723.

			a(6) = 1 since 6 = 1^2 + 1^2 + 0^2 + 2^2 with 1 + 3*1 + 5*0 = 2^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + 2^2 with 1 + 3*1 + 5*1 = 3^2.
a(15) = 2 since 15 = 2^2 + 3^2 + 1^2 + 1^2 with 2 + 3*3 + 5*1 = 4^2.
a(31) = 5 since 31 = 5^2 + 2^2 + 1^2 + 1^2 with 5 + 3*2 + 5*1 = 4^2.
a(32) = 4 since 32 = 4^2 + 0^2 + 0^2 + 4^2 with 4 + 3*0 + 5*0 = 2^2.
a(2384) = 24 since 2384 = 24^2 + 12^2 + 8^2 + 40^2 with 24 + 3*12 + 5*8 = 10^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Cf. A000118, A000290, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273107, A273108, A273110, A273134, A273278, A273294.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[Do[If[SQ[n-x^2-y^2-z^2]&&SQ[x+3y+5z],Print[n," ",x];Goto[aa]],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];Label[aa];Continue,{n,0,80}]

A273429 Number of ordered ways to write n as x^6 + y^2 + z^2 + w^2, where x,y,z,w are nonnegative integers with y <= z <= w.

Original entry on oeis.org

1, 2, 2, 2, 2, 2, 2, 1, 1, 3, 3, 2, 2, 2, 2, 1, 1, 3, 4, 3, 2, 2, 2, 1, 1, 3, 4, 4, 2, 2, 3, 1, 1, 3, 4, 3, 3, 3, 3, 2, 1, 4, 4, 2, 2, 3, 3, 1, 1, 3, 5, 5, 3, 3, 5, 3, 1, 3, 3, 3, 2, 2, 4, 2, 2, 5, 7, 5, 4, 5, 4, 1, 3, 6, 6, 6, 4, 4, 4, 1, 2

Offset: 0

Zhi-Wei Sun, May 22 2016

nonn

The author proved in arXiv:1604.06723 that for each c = 1, 4 any natural number can be written as c*x^6 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers. Thus a(n) > 0 for all n = 0,1,2,....
We note that a(n) = 1 for the following values of n not divisible by 2^6:  7, 8, 15, 16, 23, 24, 31, 32, 40, 47, 48, 56, 71, 79, 92, 112, 143, 176, 191, 240, 304, 368, 560, 624, 688, 752, 1072, 1136, 1456, 1520, 1840, 1904, 2608, 2672, 3760, 3824, 6512, 6896.
For more conjectural refinements of Lagrange's four-square theorem, one may consult the author's preprint arXiv:1604.06723.

			a(7) = 1 since 7 = 1^6 + 1^2 + 1^2 + 2^2 with 1 = 1 < 2.
a(8) = 1 since 8 = 0^6 + 0^2 + 2^2 + 2^2 with 0 < 2 = 2.
a(15) = 1 since 15 = 1^6 + 1^2 + 2^2 + 3^2 with 1 < 2 < 3.
a(16) = 1 since 16 = 0^6 + 0^2 + 0^2 + 4^2 with 0 = 0 < 4.
a(56) = 1 since 56 = 0^6 + 2^2 + 4^2 + 6^2 with 2 < 4 < 6.
a(71) = 1 since 71 = 1^6 + 3^2 + 5^2 + 6^2 with 3 < 5 < 6.
a(79) = 1 since 79 = 1^6 + 2^2 + 5^2 + 7^2 with 2 < 5 < 7.
a(92) = 1 since 92 = 1^6 + 1^2 + 3^2 + 9^2 with 1 < 3 < 9.
a(143) = 1 since 143 = 1^6 + 5^2 + 6^2 + 9^2 with 5 < 6 < 9.
a(191) = 1 since 191 = 1^6 + 3^2 + 9^2 + 10^2 with 3 < 9 < 10.
a(624) = 1 since 624 = 2^6 + 4^2 + 12^2 + 20^2 with 4 < 12 < 20.
a(2672) = 1 since 2672 = 2^6 + 4^2 + 36^2 + 36^2 with 4 < 36 = 36.
a(3760) = 1 since 3760 = 0^6 + 4^2 + 12^2 + 60^2 with 4 < 12 < 60.
a(3824) = 1 since 3824 = 2^6 + 4^2 + 12^2 + 60^2 with 4 < 12 < 60.
a(6512) = 1 since 6512 = 2^6 + 12^2 + 52^2 + 60^2 with 12 < 52 < 60.
a(6896) = 1 since 6896 = 2^6 + 36^2 + 44^2 + 60^2 with 36 < 44 < 60.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.NT], 2016-2017.
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.

Cf. A000118, A000290, A001014, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A270969, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273107, A273108, A273110, A273134, A273278, A273294, A273302, A273404, A273432, A273568.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^6-y^2-z^2],r=r+1],{x,0,n^(1/6)},{y,0,Sqrt[(n-x^6)/3]},{z,y,Sqrt[(n-x^6-y^2)/2]}];Print[n," ",r];Continue,{n,0,80}]

A273404 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with x + 24*y a square, where x,y,z,w are nonnegative integers with z <= w.

Original entry on oeis.org

1, 2, 3, 2, 2, 3, 3, 2, 1, 3, 4, 2, 1, 2, 2, 2, 2, 3, 5, 2, 3, 3, 2, 1, 1, 4, 5, 4, 2, 2, 4, 3, 3, 3, 6, 2, 6, 5, 3, 3, 3, 7, 6, 2, 2, 5, 4, 1, 2, 3, 7, 6, 8, 4, 5, 5, 2, 4, 5, 2, 3, 5, 3, 4, 2, 5, 9, 4, 5, 4, 5, 1, 3, 5, 4, 5, 5, 4, 2, 3, 3

Offset: 0

Zhi-Wei Sun, May 21 2016

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 16^k*m (k = 0,1,2,... and m = 8, 12, 23, 24, 47, 71, 168, 344, 632).

For more conjectural refinements of Lagrange's four-square theorem, one may consult arXiv:1604.06723.

			a(8) = 1 since 8 = 0^2 + 0^2 + 2^2 + 2^2 with 0 + 24*0 = 0^2.
a(12) = 1 since 12 = 1^2 + 1^2 + 1^2 + 3^2 with 1 + 24*1 = 5^2.
a(23) = 1 since 23 = 1^2 + 2^2 + 3^2 + 3^2 with 1 + 24*2 = 7^2.
a(24) = 1 since 24 = 4^2 + 0^2 + 2^2 + 2^2 with 4 + 24*0 = 2^2.
a(47) = 1 since 47 = 1^2 + 1^2 + 3^2 + 6^2 with 1 + 24*1 = 5^2.
a(71) = 1 since 71 = 1^2 + 5^2 + 3^2 + 6^2 with 1 + 24*5 = 11^2.
a(168) = 1 since 168 = 4^2 + 4^2 + 6^2 + 10^2 with 4 + 24*4 = 10^2.
a(344) = 1 since 344 = 4^2 + 0^2 + 2^2 + 18^2 with 4 + 24*0 = 2^2.
a(632) = 1 since 632 = 0^2 + 6^2 + 14^2 + 20^2 with 0 + 24*6 = 12^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Cf. A000118, A000290, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273107, A273108, A273110, A273134, A273278, A273294, A273302.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[x+24y],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[(n-x^2-y^2)/2]}];Print[n," ",r];Label[aa];Continue,{n,0,80}]

A273432 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with 2*x + y - z a nonnegative cube, where x,y,z,w are nonnegative integers with y <= z.

Original entry on oeis.org

1, 1, 2, 2, 1, 1, 4, 1, 1, 3, 1, 3, 2, 1, 3, 3, 2, 3, 5, 2, 3, 4, 6, 1, 3, 5, 1, 6, 1, 3, 7, 2, 2, 5, 6, 5, 6, 3, 6, 4, 1, 3, 4, 5, 4, 5, 7, 2, 3, 8, 6, 7, 3, 4, 8, 3, 2, 6, 3, 5, 7, 3, 8, 7, 2, 4, 10, 4, 4, 7, 9, 7, 2, 4, 2, 7, 3, 5, 11, 2, 4

Offset: 0

Zhi-Wei Sun, May 22 2016

nonn

Conjecture: (i) For each c = 1, 2, 4 and n = 0,1,2,..., we can write n as x^2 + y^2 + z^2 + w^2 with c*(2x+y-z) a nonnegative cube, where x,y,z,w are nonnegative integers with y <= z.
(ii) Each n = 0,1,2,.... can be written as x^2 + y^2 + z^2 + w^2 with x-y+z a nonnegative cube, where x,y,z,w are integers with x >= y >= 0 and x >= |z|.
The author proved in arXiv:1604.06723 that for each a = 1, 2 any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that x + y + a*z is a cube.
See also A273458 for a similar conjecture.
For more conjectural refinements of Lagrange's four-square theorem, see the author's preprint arXiv:1604.06723.

			a(1) = 1 since 1 = 0^2 + 0^2 + 0^2 + 1^2 with 0 = 0 and 2*0 + 0 - 0 = 0^3.
a(4) = 1 since 4 = 0^2 + 0^2 + 0^2 + 2^2 with 0 = 0 and 2*0 + 0 - 0 = 0^3.
a(8) = 1 since 8 = 0^2 + 2^2 + 2^2 + 0^2 with 2 = 2 and 2*0 + 2 - 2 = 0^3.
a(10) = 1 since 10 = 1^2 + 1^2 + 2^2 + 2^2 with 1 < 2 and 2*1 + 1 - 2 = 1^3.
a(13) = 1 since 13 = 2^2 + 0^2 + 3^2 + 0^2 with 0 < 3 and 2*2 + 0 - 3 = 1^3.
a(23) = 1 since 23 = 1^2 + 2^2 + 3^2 + 3^2 with 2 < 3 and 2*1 + 2 - 3 = 1^3.
a(26) = 1 since 26 = 1^2 + 3^2 + 4^2 + 0^2 with 3 < 4 and 2*1 + 3 - 4 = 1^3.
a(28) = 1 since 28 = 4^2 + 2^2 + 2^2 + 2^2 with 2 = 2 and 2*4 + 2 - 2 = 2^3.
a(40) = 1 since 40 = 4^2 + 2^2 + 2^2 + 4^2 with 2 = 2 and 2*4 + 2 - 2 = 2^3.
a(104) = 1 since 104 = 4^2 + 6^2 + 6^2 + 4^2 with 6 = 6 and 2*4 + 6 - 6 = 2^3.
a(138) = 1 since 138 = 3^2 + 5^2 + 10^2 + 2^2 with 5 < 10 and 2*3 + 5 - 10 =1^3.
a(200) = 1 since 200 = 0^2 + 10^2 + 10^2 + 0^2 with 10 = 10 and 2*0 + 10 - 10 = 0^3.
a(296) = 1 since 296 = 8^2 + 6^2 + 14^2 + 0^2 with 6 < 14 and 2*8 + 6 - 14 = 2^3.
a(328) = 1 since 328 = 0^2 + 6^2 + 6^2 + 16^2 with 6 = 6 and 2*0 + 6 - 6 = 0^3.
a(520) = 1 since 520 = 4^2 + 2^2 + 10^2 + 20^2 with 2 < 10 and 2*4 + 2 - 10 = 0^3.
a(776) = 1 since 776 = 0^2 + 10^2 + 10^2 + 24^2 with 10 = 10 and 2*0 + 10 - 10 = 0^3.
a(1832) = 1 since 1832 = 4^2 + 30^2 + 30^2 + 4^2 with 30 = 30 and 2*4 + 30 - 30 = 2^3.
a(2976) = 1 since 2976 = 20^2 + 16^2 + 48^2 + 4^2 with 16 < 48 and 2*20 + 16 - 48 = 2^3.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Cf. A000118, A000290, A000578, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A270969, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273107, A273108, A273110, A273134, A273278, A273294, A273302, A273404, A273429, A273458, A273568.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&CQ[2x+y-z],r=r+1],{x,0,n^(1/2)},{y,0,Sqrt[(n-x^2)/2]},{z,y,Min[2x+y,Sqrt[n-x^2-y^2]]}];Print[n," ",r];Continue,{n,0,80}]

A273458 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with x-y+z+w a nonnegative cube, where x,y,z,w are integers with x >= y >= 0 and x >= |z| <= |w|.

Original entry on oeis.org

1, 2, 2, 3, 2, 2, 3, 3, 2, 2, 3, 2, 1, 5, 4, 3, 2, 1, 4, 3, 3, 6, 3, 2, 5, 3, 9, 3, 1, 1, 7, 5, 3, 7, 10, 4, 6, 2, 10, 2, 6, 2, 12, 7, 2, 5, 9, 3, 3, 6, 13, 3, 8, 3, 18, 3, 8, 5, 7, 3, 3, 5, 13, 8, 5, 3, 19, 4, 7, 7, 16, 1, 11, 5, 14, 7, 2, 3, 12, 5, 4

Offset: 0

Zhi-Wei Sun, May 22 2016

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,....
In the latest version of arXiv:1605.03074, the authors showed that any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that x + y + z + w is a cube (or a square).
For more conjectural refinements of Lagrange's four-square theorem, see the author's preprint arXiv:1604.06723.

			a(12) = 1 since 12 = 3^2 + 1^2 + (-1)^2 + (-1)^2 with 3 - 1 + (-1) + (-1) = 0^3.
a(17) = 1 since 17 = 2^2 + 0^2 + 2^2 + (-3)^2 with 2 - 0 + 2 + (-3) = 1^3.
a(28) = 1 since 28 = 3^2 + 1^2 + 3^2 + 3^2 with 3 - 1 + 3 + 3 = 2^3.
a(29) = 1 since 29 = 3^2 + 0^2 + 2^2 + (-4)^2 with 3 - 0 + 2 + (-4) = 1^3.
a(71) = 1 since 71 = 5^2 + 1^2 + 3^2 + (-6)^2 with 5 - 1 + 3 + (-6) = 1^3.
a(149) = 1 since 149 = 8^2 + 0^2 + 2^2 + (-9)^2 with 8 - 0 + 2 + (-9) = 1^3.
a(188) = 1 since 188 = 13^2 + 3^2 + 1^2 + (-3)^2 with 13 - 3 + 1 + (-3) = 2^3.
a(284) = 1 since 284 = 15^2 + 5^2 + 3^2 + (-5)^2 with 15 - 5 + 3 + (-5) = 2^3.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Yu-Chen Sun and Zhi-Wei Sun, Two refinements of Lagrange's four-square theorem, arXiv:1605.03074 [math.NT], 2016.
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Cf. A000118, A000290, A000578, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A270969, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273107, A273108, A273110, A273134, A273278, A273294, A273302, A273404, A273429, A273432, A273568.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
CQ[n_]:=CQ[n]=n>=0&&IntegerQ[n^(1/3)]
Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&CQ[x-y+(-1)^j*z+(-1)^k*Sqrt[n-x^2-y^2-z^2]],r=r+1],{y,0,(n/2)^(1/2)},{x,y,Sqrt[n-y^2]},{z,0,Min[x,Sqrt[(n-x^2-y^2)/2]]},{j,0,Min[1,z]},{k,0,Min[1,Sqrt[n-x^2-y^2-z^2]]}];
Print[n," ",r];Continue,{n,0,80}]

A273568 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with w + x + 2y - 4z twice a nonnegative cube, where w is an integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 2, 2, 2, 4, 3, 3, 4, 1, 2, 2, 1, 4, 6, 2, 4, 5, 3, 5, 5, 4, 1, 4, 5, 3, 3, 3, 1, 5, 4, 4, 4, 6, 8, 5, 1, 5, 4, 3, 13, 9, 2, 6, 2, 4, 7, 9, 8, 7, 8, 5, 6, 2, 4, 5, 7, 9, 11, 5, 2, 5, 10, 6, 12, 9, 4

Offset: 0

Zhi-Wei Sun, May 25 2016

nonn

Conjecture: a(n) > 0 for all n = 0,1,2,....

For more conjectural refinements of Lagrange's four-square theorem, see the author's preprint arXiv:1604.06723.

			a(1) = 1 since 1 = 0^2 + 0^2 + 1^2 + 0^2 with 0 + 0 + 2*1 - 4*0 = 2*1^3.
a(3) = 1 since 3 = (-1)^2 + 1^2 + 1^2 + 0^2 with (-1) + 1 + 2*1 - 4*0 = 2*1^3.
a(13) = 1 since 13 = (-2)^2 + 2^2 + 2^2 + 1^2 with (-2) + 2 + 2*2 - 4*1 = 2*0^3.
a(16) = 1 since 16 = 2^2 + 2^2 + 2^2 + 2^2 with 2 + 2 + 2*2 - 4*2 = 2*0^3.
a(26) = 1 since 26 = 3^2 + 3^2 + 2^2 + 2^2 with 3 + 3 + 2*2 - 4*2 = 2*1^3.
a(32) = 1 since 32 = (-4)^2 + 4^2 + 0^2 + 0^2 with (-4) + 4 + 2*0 - 4*0 = 2*0^3.
a(40) = 1 since 40 = (-2)^2 + 4^2 + 4^2 + 2^2 with (-2) + 4 + 2*4 - 4*2 = 2*1^3.
a(218) = 1 since 218 = (-6)^2 + 6^2 + 11^2 + 5^2 with (-6) + 6 + 2*11 - 4*5 = 2*1^3.
a(416) = 1 since 416 = (-4)^2 + 20^2 + 0^2 + 0^2 with (-4) + 20 + 2*0 - 4*0 = 2*2^3.
a(544) = 1 since 544 = (-4)^2 + 20^2 + 8^2 + 8^2 with (-4) + 20 + 2*8 - 4*8 = 2*0^3.
a(800) = 1 since 800 = (-20)^2 + 20^2 + 0^2 + 0^2 with (-20) + 20 + 2*0 - 4*0 = 2*0^3.
a(1184) = 1 since 1184 = (-28)^2 + 12^2 + 16^2 + 0^2 with (-28) + 12 + 2*16 - 4*0 = 2*2^3.
a(2080) = 1 since 2080 = (-20)^2 + 20^2 + 32^2 + 16^2 with (-20) + 20 + 2*32 - 4*16 = 2*0^3.
a(6304) = 1 since 6304 = (-36)^2 + 36^2 + 56^2 + 24^2 with (-36) + 36 + 2*56 - 4*24 = 2*2^3.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Cf. A000118, A000290, A000578, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A270969, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273107, A273108, A273110, A273134, A273278, A273294, A273302, A273404, A273429, A273432, A273458.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
CQ[n_]:=CQ[n]=n>=0&&IntegerQ[n^(1/3)]
Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&CQ[(x+2y-4z+(-1)^k*Sqrt[n-x^2-y^2-z^2])/2],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]},{k,0,Min[1,n-x^2-y^2-z^2]}];Print[n," ",r];Continue,{n,0,70}]

A340274 Number of ways to write n as x + y + z with x, y, z positive integers such that 3x^2y^2 + 5y^2z^2 + 8z^2x^2 is a square.

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 2, 4, 2, 5, 2, 1, 2, 2, 2, 2, 4, 2, 4, 3, 3, 3, 4, 5, 2, 3, 5, 5, 4, 4, 2, 4, 4, 5, 3, 4, 3, 6, 3, 2, 5, 2, 2, 7, 7, 1, 3, 6, 4, 4, 3, 3, 6, 2, 5, 5, 3, 6, 5, 4, 6, 6, 6, 3, 6, 6, 4, 5, 6, 2, 6, 3, 5, 4, 5, 3, 5, 12, 4, 4, 5, 1, 6, 6, 7, 9, 3, 3, 6, 5, 6, 7, 4

Offset: 1

Zhi-Wei Sun, Apr 24 2021

nonn

Conjecture 1: a(n) > 0 for all n > 2.
We have verified a(n) > 0 for all n = 3..10000. The conjecture holds if a(p) > 0 for every odd prime p. For any n > 0 we have a(3*n) > 0, since 3*n = n + n + n and 3 + 5 + 8 = 4^2.
It seems that a(n) = 1 only for n = 3..8, 10, 11, 19, 53, 89, 127, 178, 257, 461.
See also A343862 for similar conjectures.
Conjecture 1 holds for all n < 2^15. Note a(1823) = 1. - Martin Ehrenstein, May 03 2021

			a(4) = 1 with 4 = 2 + 1 + 1 and 3*2^2*1^2 + 5*1^2*1^2 + 8*1^2*2^2 = 7^2.
a(19) = 1 with 19 = 9 + 9 + 1 and 3*9^2*9^2 + 5*9^2*1^2 + 8*1^2*9^2 = 144^2.
a(53) = 1 with 53 = 23 + 7 + 23 and 3*23^2*7^2 + 5*7^2*23^2 + 8*23^2*23^2 = 1564^2.
a(89) = 1 with 89 = 2 + 58 + 29 and 3*2^2*58^2 + 5*58^2*29^2 + 8*29^2*2^2 = 3770^2.
a(257) = 1 with 257 = 11 + 164 + 82 and 3*11^2*164^2 + 5*164^2*82^2 + 8*82^2*11^2 = 30340^2.
a(461) = 1 with 461 = 186 + 165 + 110 and 3*186^2*165^2 + 5*165^2*110^2 + 8*110^2*186^2 = 88440^2.

Martin Ehrenstein, Table of n, a(n) for n = 1..32767 (first 1500 terms from Zhi-Wei Sun)
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175 (2017), 167-190. See also arXiv version, arXiv:1604.06723 [math.NT], 2016-2017.

Cf. A000041, A000290, A230121, A230747, A231168, A262357, A271719, A273278, A343862.

SQ[n_]:=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[3x^2*y^2+(n-x-y)^2*(5*y^2+8*x^2)],r=r+1],{x,1,n-2},{y,1,n-1-x}];tab=Append[tab,r],{n,1,100}];Print[tab]

A343862 Number of ways to write n as x + y + z with x, y, z positive integers such that x^2y^2 + 5y^2z^2 + 10z^2*x^2 is a square.

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 2, 1, 1, 4, 2, 4, 2, 2, 5, 5, 3, 3, 2, 6, 4, 3, 3, 6, 6, 5, 2, 6, 4, 10, 3, 6, 4, 6, 6, 8, 5, 6, 4, 9, 7, 6, 3, 7, 9, 5, 5, 15, 5, 12, 11, 10, 5, 6, 7, 10, 8, 9, 7, 15, 7, 6, 7, 10, 10, 7, 9, 10, 10, 12, 4, 15, 9, 9, 11, 9, 7, 12, 11, 15, 8, 9, 7, 12, 10, 3, 9, 11, 11, 19, 12, 12, 9, 10, 6, 23, 11, 6, 10, 18

Offset: 1

Zhi-Wei Sun, May 02 2021

nonn

Conjecture 1: a(n) > 0 for all n > 2.
We have verified a(n) > 0 for all n = 3..10000. Conjecture 1 holds if a(p) > 0 for each odd prime p. For any n > 0 we have a(3*n) > 0 since 3*n = n + n + n and 1 + 5 + 10 = 4^2.
See also A340274 for a similar conjecture.
Conjecture 2: There are infinitely many triples (a,b,c) of positive integers such that each n = 3,4,... can be written as x + y + z with x,y,z positive integers and a*x^2*y^2 + b*y^2*z^2 + c*z^2*x^2 a square.
Such triple candidates include (21,19,9), (23,17,9), (24,16,9), (25,14,10), (29,19,16), (33,27,21), (35,9,5), (37,32,31) etc.
Conjecture 1 holds for all n < 2^15. - Martin Ehrenstein, May 02 2021

			a(4) = 1 with 4 = 2 + 1 + 1 and 2^2*1^2 + 5*1^2*1^2 + 10*1^2*2^2 = 7^2.
a(5) = 1 with 5 = 1 + 3 + 1 and 1^2*3^2 + 5*3^2*1^2 + 10*1^1*1^2 = 8^2.
a(8) = 1 with 8 = 4 + 2 + 2 and 4^2*2^2 + 5*2^2*2^2 + 10*2^2*4^2 = 28^2.
a(9) = 1 with 9 = 3 + 3 + 3 and 3^2*3^2 + 5*3^2*3^2 + 10*3^2*3^2 = 36^2.
a(19) = 2. We have 19 = 4 + 5 + 10 with 4^2*5^2 + 5*5^2*10^2 + 10*10^2*4^2 = 170^2, and 19 = 4 + 13 + 2 with 4^2*13^2 + 5*13^2*2^2 + 10*2^2*4^2 = 82^2.

Martin Ehrenstein, Table of n, a(n) for n = 1..32767 (first 1500 terms from Zhi-Wei Sun)
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. See also arXiv:1604.06723 [math.NT].

Cf. A000041, A000290, A230121, A230747, A231168, A262357, A271719, A273278, A340274.

SQ[n_]:=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[x^2*y^2+(n-x-y)^2*(5*y^2+10*x^2)],r=r+1],{x,1,n-2},{y,1,n-1-x}];tab=Append[tab,r],{n,1,100}];Print[tab]

A273616 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with (3x^2+13y^2)*z a square, where x,y,z,w are nonnegative integers.

Original entry on oeis.org

1, 4, 4, 2, 5, 8, 4, 2, 4, 8, 11, 4, 2, 10, 8, 1, 4, 12, 10, 8, 9, 8, 9, 1, 4, 17, 16, 6, 3, 16, 8, 1, 4, 8, 18, 10, 8, 12, 13, 2, 10, 18, 9, 8, 5, 17, 11, 3, 2, 15, 22, 7, 13, 15, 17, 4, 6, 10, 11, 14, 2, 18, 17, 1, 5, 23, 13, 9, 13, 14, 14, 1, 8, 16, 26, 8, 4, 16, 7, 1, 8

Offset: 0

Zhi-Wei Sun, May 26 2016

nonn

Conjecture: For each ordered pair (a,b) = (3,13), (5,11), (15,57), (15,165), (138,150), any natural number can be written as x^2 + y^2 + z^2 + w^2 with (a*x^2+b*y^2)*z a square, where x,y,z,w are nonnegative integers.

For more conjectural refinements of Lagrange's four-square theorem, see the author's preprint arXiv:1604.06723.

			a(15) = 1 since 15 = 2^2 + 1^2 + 1^2 + 3^2 with (3*2^2+13*1^2)*1 = 5^2.
a(23) = 1 since 23 = 3^2 + 3^2 + 1^2 + 2^2 with (3*3^2+13*3^2)*1 = 12^2.
a(31) = 1 since 31 = 2^2 + 1^2 + 1^2 + 5^2 with (3*2^2+13*1^2)*1 = 5^2.
a(63) = 1 since 63 = 6^2 + 1^2 + 1^2 + 5^2 with (3*6^2+13*1^2)*1 = 11^2.
a(71) = 1 since 71 = 6^2 + 3^2 + 1^2 + 5^2 with (3*6^2+13*3^2)*1 = 15^2.
a(79) = 1 since 79 = 5^2 + 3^2 + 3^2 + 6^2 with (3*5^2+13*3^2)*3 = 24^2.
a(223) = 1 since 223 = 2^2 + 13^2 + 1^2 + 7^2 with (3*2^2+13*13^2)*1 = 47^2.
a(303) = 1 since 303 = 2^2 + 13^2 + 9^2 + 7^2 with (3*2^2+13*13^2)*9 = 141^2.
a(2703) = 1 since 2703 = 15^2 + 25^2 + 22^2 + 37^2 with (3*15^2+13*25^2)*22 = 440^2.

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Cf. A000118, A000290, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A270969, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977, A273021, A273107, A273108, A273110, A273134, A273278, A273294, A273302, A273404, A273429, A273432, A273458, A273568.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[(3x^2+13y^2)z],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];Print[n," ",r];Label[aa];Continue,{n,0,80}]

cp's OEIS Frontend

A273294 Least nonnegative integer m such that there are nonnegative integers x,y,z,w for which x^2 + y^2 + z^2 + w^2 = n and x + 3*y + 5*z = m^2.

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A273302 Least nonnegative integer x such that n = x^2 + y^2 + z^2 + w^2 for some nonnegative integer y,z,w with x + 3*y + 5*z a square.

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A273429 Number of ordered ways to write n as x^6 + y^2 + z^2 + w^2, where x,y,z,w are nonnegative integers with y <= z <= w.

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A273404 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with x + 24*y a square, where x,y,z,w are nonnegative integers with z <= w.

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A273432 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with 2*x + y - z a nonnegative cube, where x,y,z,w are nonnegative integers with y <= z.

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A273458 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with x-y+z+w a nonnegative cube, where x,y,z,w are integers with x >= y >= 0 and x >= |z| <= |w|.

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A273568 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with w + x + 2*y - 4*z twice a nonnegative cube, where w is an integer and x,y,z are nonnegative integers.

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A340274 Number of ways to write n as x + y + z with x, y, z positive integers such that 3*x^2*y^2 + 5*y^2*z^2 + 8*z^2*x^2 is a square.

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A273294 Least nonnegative integer m such that there are nonnegative integers x,y,z,w for which x^2 + y^2 + z^2 + w^2 = n and x + 3y + 5z = m^2.

A273302 Least nonnegative integer x such that n = x^2 + y^2 + z^2 + w^2 for some nonnegative integer y,z,w with x + 3y + 5z a square.

A273568 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with w + x + 2y - 4z twice a nonnegative cube, where w is an integer and x,y,z are nonnegative integers.

A340274 Number of ways to write n as x + y + z with x, y, z positive integers such that 3x^2y^2 + 5y^2z^2 + 8z^2x^2 is a square.

A343862 Number of ways to write n as x + y + z with x, y, z positive integers such that x^2y^2 + 5y^2z^2 + 10z^2*x^2 is a square.

A273616 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with (3x^2+13y^2)*z a square, where x,y,z,w are nonnegative integers.